Data transmission method and apparatus in wireless local area network

ABSTRACT

An HE-LTF transmission method is provided, including: determining, based on a total number NSTS of space-time streams, a number NHELTF of OFDM symbols included in an HE-LTF field; determining a HE-LTF sequence in frequency domain according to a transmission bandwidth and a mode of the HE-LTF field, where the HE-LTF sequence in frequency domain includes but is not limited to a mode of the HE-LTF field sequence that is in a 1× mode and that is mentioned in implementations; and sending a time-domain signal according to the number NHELTF of OFDM symbols and the determined HE-LTF sequence in frequency domain. In the foregoing solution, a PAPR value is relatively low.

This application is a continuation of International Application No.PCT/CN2016/106941, filed on Nov. 23, 2016 which claims priority toChinese Patent Application No. 201510854631.4, filed with the ChinesePatent Office on Nov. 30, 2015 and entitled “DATA TRANSMISSION METHODAND APPARATUS IN WIRELESS LOCAL AREA NETWORK”, and claims priority toChinese Patent Application No. 201510823977.8, filed with the ChinesePatent Office on Nov. 23, 2015 and entitled “DATA TRANSMISSION METHODAND APPARATUS IN WIRELESS LOCAL AREA NETWORK”, which are incorporatedherein by reference in their entireties.

TECHNICAL FIELD

The present invention relates to the communications field, and inparticular, to a service message construction method and apparatus.

BACKGROUND

A wireless local area network (WLAN) is a data transmission system, andreplaces, by using a radio frequency (RF) technology, a legacy localarea network including a twisted-pair copper wire, so that a user cantransmit information via the wireless local area network by using asimple access architecture. Development and application of a WLANtechnology have greatly changed people's communication manner andworking manner, and bring unprecedented convenience to people. Wideapplication of intelligent terminals is accompanied by people's growingrequirements for data network traffic. Development of the WLAN dependson standard formulation, popularization, and application. The IEEE802.11 family is primary standards, and mainly includes 802.11,802.11b/g/a, 802.11n, and 802.11ac. In all standards except the 802.11and the 802.11b, an orthogonal frequency division multiplexing (OFDM)technology is used as a core technology at a physical layer.

Channel estimation is a process of estimating, according to a receivesignal and by a specific criterion, a parameter of a channel throughwhich a transmit signal passes. Performance of a wireless communicationssystem is affected by a wireless channel to a great extent, such asshadow fading and frequency selective fading. Consequently, atransmission path between a transmitter and a receiver is extremelycomplex. Unlike a wired channel that is fixed and predictable, thewireless channel is characterized by high randomness. A channel needs tobe estimated in coherent detection of an OFDM system, and channelestimation precision directly affects performance of the entire system.

SUMMARY

To reduce a PAPR of a wireless local area network, the presentembodiment provides an HE-LTF transmission method, including:determining, based on a total number of space-time streams, N_(STS), anumber of OFDM symbols of an HE-LTF field, N_(HELTF); determining anHE-LTF sequence in frequency domain according to a transmissionbandwidth and a mode of the HE-LTF field, where the HE-LTF sequence infrequency domain includes but is not limited to the HE-LTF sequence of1× mode and that is mentioned in implementations; and sending atime-domain signal(s) according to the number N_(HELTF) of OFDM symbolsand the determined HE-LTF sequence in frequency domain.

In addition, correspondingly, an HE-LTF transmission method is provided,including: obtaining a transmission bandwidth BW, a total number ofspace-time streams, N_(STS), and an mode of an HE-LTF field according toinformation carried in a signal field in a preamble; determining, basedon the total number of space-time streams, N_(STS), a number of OFDMsymbols included in an HE-LTF field, N_(HELTF); determining acorresponding HE-LTF sequence in frequency domain according to thetransmission bandwidth and the HE-LTF field mode, where the HE-LTFsequence in frequency domain includes but is not limited to the HE-LTFsequenceof 1× mode and that is mentioned in implementations; andobtaining a channel estimation value of a corresponding subcarrierlocation according to the received HE-LTF field and the determinedsequence in frequency domain.

By means of simulation and comparison, the HE-LTF sequence of the 1×mode in the embodiment is used, so that a system has an extremely lowPAPR value.

BRIEF DESCRIPTION OF DRAWINGS

To describe the technical solutions in the embodiments of the presentinvention more clearly, the following briefly describes the accompanyingdrawings required for describing the embodiments or the prior art.Apparently, the accompanying drawings in the following description showsome embodiments of the present invention, and a person of ordinaryskill in the art may still derive other drawings from these accompanyingdrawings without creative efforts.

FIG. 1 is a simple schematic diagram of a format of an HE PPDU;

FIG. 2 is a schematic diagram of a tone plan in a 20 MHz bandwidth;

FIG. 3A and FIG. 3B are a schematic diagram of a tone plan in a 40 MHzbandwidth;

FIG. 4A and FIG. 4B are a schematic diagram of a tone plan over an 80MHz bandwidth;

FIG. 5 is a schematic diagram of simple comparison of 1×, 2×, and 4×OFDMsymbols in a frequency domain;

FIG. 6 is a simple schematic diagram of a system architecture in anembodiment;

FIG. 7 is a simple schematic diagram of generating and sending an HE-LTFfield during sending of an SU or a downlink DL MU MIMO data packet;

FIG. 8 is a simple schematic diagram of generating and sending an HE-LTFfield during sending of a UL MU MIMO data packet;

FIG. 9A, FIG. 9B, and FIG. 9C are block diagrams of a transmit end of adata transmission apparatus on a subcarrier location B of a 20 M1×HE-LTF in an embodiment;

FIG. 10 is a block diagram of a receive end of a data transmissionapparatus on a subcarrier location B of a 20 M 1×HE-LTF in anembodiment; and

FIG. 11 is a simple schematic diagram of a data transmission apparatusin an embodiment.

DESCRIPTION OF EMBODIMENTS

To make the objectives, technical solutions, and advantages of theembodiments of the present invention clearer, the following clearlydescribes the technical solutions in the embodiments of the presentinvention with reference to the accompanying drawings in the embodimentsof the present invention. Apparently, the described embodiments are somebut not all of the embodiments of the present invention. All otherembodiments obtained by a person of ordinary skill in the art based onthe embodiments of the present invention without creative efforts shallfall within the protection scope of the present invention.

Solutions of the embodiments of the present invention may be applicableto a WLAN network system. FIG. 6 is a schematic diagram of a scenario towhich a pilot transmission method in a wireless local area network isapplicable according to Embodiment 1 of the present invention. As shownin FIG. 6, the WLAN network system may include one access point 101 andat least two stations 102.

An access point (AP) may also be referred to as a wireless access point,a bridge, a hotspot, or the like, and may access a server or acommunications network.

The station (STA) may also be referred to as user equipment, and may bea wireless sensor, a wireless communications terminal, or a mobileterminal, such as a mobile phone (or referred to as a “cellular” phone)that supports a Wi-Fi communication function and a computer with awireless communication function. For example, the station may be aportable, pocket-sized, handheld, computer built-in, wearable, orin-vehicle wireless communications apparatus that supports a Wi-Ficommunication function, which exchanges communication data such as avoice or data with a radio access network. A person skilled in the artlearns that some communications devices may have functions of both theforegoing access point and the foregoing station, and no limitation isimposed herein.

A common point of the foregoing WLAN standards that use an OFDMtechnology as a core is that a long training field (LTF) that can beused for channel estimation is stipulated at a physical layer. Forexample, FIG. 1 shows a format that is of a high efficiency (HE)physical protocol data unit (PPDU) and that is stipulated in the 802.11ax standard. An HE-LTF field is a high efficiency long training fieldused for channel estimation of a data part. This field may include oneor more HE-LTF elements, and each element is an OFDM symbol.

To improve a system throughput rate, the OFDMAtechnology is introducedinto the 802.11ax standard. A corresponding subcarrier spacing at aphysical layer reduces from existing Δ_(F) ^(1×)=20 MHz/64=312.5 kHz toΔ_(F) ^(4×)=20 MHz/256=78.125 kHz, and a Fourier transform period of anOFDM symbol of a data part at the physical layer also changes fromT_(DFT) ^(1×)=1/Δ_(F) ^(1×)=3.2 us to T_(DFT) ^(4×)=1/Δ_(F) ^(4×)=12.8us. Sometimes, the subcarrier spacing is changed to Δ_(F) ^(2×)=20MHz/128=156.25 kHz. Formats of the foregoing different OFDM symbols arerespectively referred to as a 4× mode, a 2× mode, and a 1× mode forshort.

As the 802.11ax standard gradually evolves, FIG. 2 to FIG. 4A and FIG.4B show tone plans in a 20 MHz bandwidth, a 40 MHz bandwidth, an 80 MHzbandwidth, and a 160/80+80 MHz bandwidth. Tone plans in a left 80 MHzbandwidth and a right 80 MHz bandwidth of the 160/80+80 MHz are the sameas a tone plan in the 80 MHz bandwidth. The tone plan shows a possiblelocation and size of a resource unit during scheduling.

In the 20 MHz bandwidth, pilot subcarrier locations of 242 RUs (resourceunit) are ±22, ±48, ±90, and ±116. In the 40 MHz bandwidth, pilotsubcarrier locations of 484 RUs are ±10, ±36, ±78, ±104, ±144, ±170,±212, and ±238. In the 80 MHz bandwidth, pilot subcarrier locations of996 RUs are ±24, ±92, ±158, ±226, ±266, ±334, ±400, and ±468.

To further improve system efficiency in different scenarios, the HE-LTFfield needs to support OFDM symbols in the foregoing 4× mode, 2× mode,and 1× mode.

As shown in FIG. 5, a 20 MHz bandwidth is used as an example. Whensubcarrier locations are marked as −128, −127, . . . , −2, −1, 0, 1, 2,. . . , and 127, in a 4× mode, subcarriers in an HE-LTF element thatcarry a long training sequence are located in locations (indexes) −122,−121, . . . , −3, −2, 2, 3, . . . , 121, and 122, remaining subcarriersare empty subcarriers, and a subcarrier spacing is Δ_(F) ^(4×)=20MHz/256=78.125 kHz.

In a 2× mode, subcarriers in an HE-LTF element that carry a longtraining sequence are located in −122, −120, . . . , −4, −2, 2, 4, . . ., 120, and 122, and remaining subcarriers are empty subcarriers.Equivalently, subcarrier locations may be marked as −64, −63, . . . ,−2, −1, 0, 1, 2, . . . , and 63. In this case, the subcarriers in theHE-LTF element in the 2× mode that carry a long training sequence arelocated in −61, −60, . . . , −2, −1, 1, 2, . . . , 60, and 61, and theremaining subcarriers are empty subcarriers. In this case, a subcarrierspacing is Δ_(F) ^(2×)=20 MHz/128=156.25 kHz.

Similarly, in a ix mode, subcarriers in an HE-LTF element that carry along training sequence are located in −120, −116, . . . , −8, −4, 4, 8,. . . , 116, and 120, and remaining subcarriers are empty subcarriers.Equivalently, subcarrier locations may be marked as −32, −31, . . . ,−2, −1, 0, 1, 2, . . . , and 31. In this case, in the ix mode, thesubcarriers in the HE-LTF element that carry a long training sequenceare located in −30, −29, . . . , −2, −1, 1, 2, . . . , 29, and 30, andthe remaining subcarriers are empty subcarriers. In this case, asubcarrier spacing is Δ_(F) ^(1×)=20 MHz/64=312.5 kHz.

Currently, only a 4×HE-LTF sequence and a 2×HE-LTF sequence aredetermined, and a 1×HE-LTF sequence has not been determined. It is stillopen in terms of how to define the 1×HE-LTF sequence.

In the 11n standard and the 11ac standard, a subcarrier spacing is Δ_(F)^(1×), and a 20 MHz HT/VHT LTF sequence is defined as follows:

BB_LTF_L={+1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1}

BB_LTF_R={+1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1}

LTF_(left)={BB_LTF_L, BB_LTF_L}={+1, +1, −1, −1, +1, +1, −1, +1, −1, +1,+1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1}

LTF_(right)={BB_LTF_R, −1×BB_LTF_R}={+1, −1, −1, +1, +1, −1, +1, −1, +1,−1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1}

VHT-LTF₅₆(−28:28)={+1, +1, LTF_(left), 0, LTF_(right), −1, −1}

However, subcarriers in a 1×HE-LTF element that carry a long trainingsequence are located in 60 non-empty subcarriers in total: −30, −29, . .. , −2, −1, 1, 2, . . . , 29, and 30. LTF sequences in the existing 11nand 11ac standards cannot be directly used. A similar problem alsoexists in other bandwidths.

A 1×HE-LTF is mainly applied to an OFDM communication scenario ratherthan an OFDMA communication scenario. PAPR values of HE-LTF symbolsgenerated when different RUs are scheduled do not need to be considered,and only a PAPR value of an HE-LTF symbol during OFDM transmission ineach full bandwidth needs to be considered, for example, the 242 RU in20 MHz, the 484 RU in 40 MHz, or the 996 RU in 80 MHz. Therefore, in thepresent embodiment, based on sequences BB_LTF_L, BB_LTF_R, LTF_(left),and LTF_(right) that are characterized by an excellent PAPR, a series ofextension operations are performed on the sequences, to obtain new1×HE-LTF sequences characterized by a low PAPR in different bandwidths.The following sequence −1×BB_LTF_L indicates that polarity of each valuein a sequence BB_LTF_L is reversed, that is, 1 is changed to −1, and −1is changed to 1. The same is true for −1×BB_LTF_R, −1 LTF_(left), −1LTF_(right), and the like.

The present embodiment provides a method for sending an SU (single user,single user) data packet or a DL-MU-MIMO (Down Link Multi-user Multiplein Multiple out, downlink multi-user multiple-input multiple-output)data packet by a transmit end, including a process of generating anHE-LTF field.

A number of OFDM symbols of an HE-LTF field, N_(HELTF), is determined,on the basis of a total number of space-time streams, N_(STS).

A HE-LTF sequence in frequency domain is determined according to atransmission bandwidth and a mode of the HE-LTF field. The HE-LTFsequence in frequency domain includes but is not limited to sequencesmentioned in implementations.

Time-domain signals are sent according to the number of OFDM symbolsN_(HELTF) and the determined HE-LTF sequence in frequency domain.

Specifically, at a transmit end, the following steps are performed:

101. Determine, based on a total number of space-time streams, N_(STS),a number of OFDM symbols of an HE-LTF field, N_(HELTF). A specificcorrespondence is provided in the following Table 1.

TABLE 1 N_(STS) N_(HELTF) 1 1 2 2 3 4 4 4 5 6 6 6 7 8 8 8

102. Determine a HE-LTF sequence in frequency domain according to atransmission bandwidth and an mode of the HE-LTF field. For example,when the transmission bandwidth is BW=20 MHz, and the mode of the HE-LTFfield is a 1× mode, the HE-LTF sequence in frequency domain iscorrespondingly an HE-LTF sequence in Embodiment 1.

103. If N_(HELTF)>1, determine that a used orthogonal mapping matrix Aincludes N_(HELTF) rows and N_(HELTF) columns. Specially, whenN_(HELTF)=1, the orthogonal mapping matrix A is degenerated into 1. Avalue of a sequence carried by a subcarrier of each OFDM symbol in theHE-LTF field is multiplied by the orthogonal mapping matrix A in thefollowing manner. As shown in FIG. 7, when the number of space-timestreams is N_(STS), a value of a sequence carried by a k^(th) subcarrierof an n^(th) OFDM symbol of an i^(th) spatial stream in the HE-LTF fieldis multiplied by [A_(HELTF) ^(k)]_(i,n), where i=1, . . . N_(STS), n=1,. . . N_(HELTF).

The orthogonal mapping matrix A is defined as follows:

$A_{HELTF}^{k} = \left\{ {\begin{matrix}{R,} & {{{if}\mspace{14mu} k} \in K_{pilot}} \\{P,} & {otherwise}\end{matrix},} \right.$

where

K_(Pilot) is a pilot subcarrier set, a matrix P is defined as

$P = \left\{ {{{\begin{matrix}{P_{4 \times 4},{N_{STS} \leq 4}} & \; \\{P_{6 \times 6},{N_{STS} = 5},6,} & {{w = {\exp \left( {{- j}\; 2{\pi/6}} \right)}},} \\{P_{8 \times 8},{N_{STS} = 7},8} & \;\end{matrix}P_{4 \times 4}} = {{\begin{bmatrix}1 & {- 1} & 1 & 1 \\1 & 1 & {- 1} & 1 \\1 & 1 & 1 & {- 1} \\{- 1} & 1 & 1 & 1\end{bmatrix}P_{6 \times 6}} = {{\begin{bmatrix}1 & {- 1} & 1 & 1 & 1 & {- 1} \\1 & {- w^{1}} & w^{2} & w^{3} & w^{4} & {- w^{5}} \\1 & {- w^{2}} & w^{4} & w^{6} & w^{8} & {- w^{10}} \\1 & {- w^{3}} & w^{6} & w^{9} & w^{12} & {- w^{15}} \\1 & {- w^{4}} & w^{8} & w^{12} & w^{16} & {- w^{20}} \\1 & {- w^{5}} & w^{10} & w^{15} & w^{20} & {- w^{25}}\end{bmatrix}P_{8 \times 8}} = \begin{bmatrix}P_{4 \times 4} & P_{4 \times 4} \\P_{4 \times 4} & {- P_{4 \times 4}}\end{bmatrix}}}},} \right.$

and

a matrix R is defined as [R]_(m,n)=[P]_(l,n).

104. Perform different cyclic shift delay on each space-time stream inthe HE-LTF field. A cyclic shift value corresponding to each space-timestream is shown in the following Table 2.

TABLE 2 T_(CS) values for the VHT modulated fields of a PPDU Totalnumberof space-time streams Cyclic shift for space-time stream i (ns)(N_(STS)) 1 2 3 4 5 6 7 8 1 0 — — — — — — — 2 0 −400 — — — — — — 3 0−400 −200 — — — — — 4 0 −400 −200 −600 — — — — 5 0 −400 −200 −600 −350 —— — 6 0 −400 −200 −600 −350 −650 — — 7 0 −400 −200 −600 −350 −650 −100 —8 0 −400 −200 −600 −350 −650 −100 −750

105. Map the space-time stream(s) in the HE-LTF field to transmitchain(s). If a total number of transmit chains is N_(TX) and the totalnumber of space-time streams is N_(STS) an antenna mapping matrix

_(k) of a k^(th) subcarrier includes N_(TX) rows and N_(STS) columns.The matrix

_(k) may be a matrix defined in chapter 20.3.11.11.2 in the 802.11nstandard.

106. Obtain time-domain signal(s) of the HE-LTF field by means ofinverse discrete Fourier transform, and send the time-domain signal (3).

At a receive end, the following steps are performed:

201. Obtain a transmission bandwidth BW, a total number of space-timestreams, N_(STS), and a mode of an HE-LTF field, according toinformation carried in a signal field in a preamble. The HE-LTF fieldmode is also referred to as an HE-LTF symbol mode, that is, theforegoing 1× mode, 2× mode, or 4× mode.

202. Determine, based on the total number of space-time streams,N_(STS), a number of OFDM symbols of the HE-LTF field, N_(HELTF).

203. Determine a corresponding HE-LTF sequence in frequency domainaccording to the transmission bandwidth and the mode of the HE-LTFfield; and obtain a channel estimation value of a correspondingsubcarrier location, based on the received HE-LTF field and thedetermined HE-LTF sequence in frequency domain.

In another example, there is a difference between a manner of generatingthe HE-LTF field during sending a UL-MU-MIMO (Up Link Multi-userMultiple in Multiple out, uplink multi-user multiple-inputmultiple-output) data packet, and a manner of generating an HE-LTF fieldduring sending an SU data packet or a DL-MU-MIMO data packet; thedifference lies in that: before a non-AP station sends the UL-MU-MIMOdata packet, an AP needs to indicate uplink scheduling information byusing a trigger frame, and the uplink scheduling information includesidentifiers of scheduled stations, a transmission bandwidth, a totalnumber of space-time streams (or a number of HE-LTF symbols), and asequence number of a spatial stream allocated to the scheduled stations.

At a transmit end, the following steps are performed:

301. Determine, a number of OFDM symbols of an HE-LTF field, N_(HELTF),based on a total number of space-time streams, N_(STS). If thescheduling information includes information of the number of HE-LTFsymbols, this step may be omitted.

302. Determine a HE-LTF sequence in frequency domain according to atransmission bandwidth and a mode of the HE-LTF field. For example, whenthe transmission bandwidth is BW=40 MHz and the mode of the HE-LTF fieldis a 1× mode, the HE-LTF sequence in frequency domain is correspondinglyan HE-LTF sequence in the following Embodiment 2.

303. Perform masking (that is, exclusive OR) processing on the HE-LTFsequence by using a row sequence corresponding to a sequence number of aspatial stream allocated to the transmit end (that is, a scheduled user)in an 8×8 matrix P. For example, when an initial HE-LTF sequence is {L₁,L₂, . . . , L_(m)}, and the sequence number of the spatial streamallocated to the transmit end is {i₁, i₂, i₃}, an {i₁, i₂, i₃}^(th) rowin the 8×8 matrix P is selected for a mask sequence. In this case, amasked HE-LTF sequence of an i_(l) ^(th) spatial stream is:

HELTF_(k) ^(i) ¹ ={L ₁ P _(i) ₁ _(,1) ,L ₂ P _(i) ₁ _(,2) , . . . ,L ₈ P_(i) ₁ _(,8) ,L ₉ P _(i) ₁ _(,1) ,L ₁₀ P _(i) ₁ _(,2) , . . . ,L ₁₆ P_(i) ₁ _(,8) ,L ₁₇ P _(i) ₁ _(,1) , . . . ,L _(m) P _(i) ₁_(((m−1)mod8)+1)},

where

mod indicates a modulo operation. Likewise, HELFT_(k) ^(i) ² andHELTF_(k) ^(i) ³ may be obtained.

304. Determine that a used orthogonal mapping matrix A includesN^(HELTF) rows and N_(HELTF) columns. A value of a sequence carried by asubcarrier of each OFDM symbol in the HE-LTF field is multiplied by theorthogonal mapping matrix A in the following manner.

For example, as shown in FIG. 8, when the sequence number of the spatialstream allocated to the transmit end (that is, the scheduled user) is{i₁, i₂, i₃}, a value HELTF′_(k) of a sequence carried by a k^(th)subcarrier of an n^(th) OFDM symbol in the HE-LTF field is multiplied by[A_(HELTF) ^(k)], where i=i₁, i₂, i₃, and n=1, . . . N_(HELTF).Optionally, the matrix A in FIG. 7 may be replaced with a matrix P.

Remaining steps are similar to those in the foregoing example, anddetails are not described herein.

At a receive end, since it's a UL-MU-MIMO transmission and an AP knowsrelated scheduling information, a channel estimation algorithm may bedirectly performed.

401. Obtain a channel estimation value of a corresponding subcarrierlocation, based on a received HE-LTF field and a known frequency domainsequence.

It can be understood that a CSD value, the matrix Q, and the like in theforegoing example are only examples, and other values may be selected.This is not limited in the embodiment.

Preferable HE-LTF sequences in a 1× mode in various bandwidths aredescribed below by using examples.

Embodiment 1

Scenario: a subcarrier location A of a 1×HE-LTF in a 20 MHz bandwidth.

For example, additional eight subcarrier values are added based on twosequences BB_LTF_L and two sequences BB_LTF_R, so as to generate a1×HE-LTF sequence. To ensure simple implementation, the eight subcarriervalues are selected from {1, −1}.

An optimal sequence is: HE-LTF₆₀(−120:4:120)={BB_LTF_L, +1, −1,−1×BB_LTF_L, −1, −1, 0, +1, +1, BB_LTF_R, −1, −1, BB_LTF_R}, or may berepresented as

HE-LTF 60(−120:4:120) = {+1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, 0, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein, asdescribed above, −120:4:120 represents −120, −116, . . . , −8, −4, 0, 4,8, . . . , 116, and 120. In this case, corresponding pilot subcarrierlocations are ±48 and ±116, that is, there are four pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.1121 dB.

Referring to Table 3, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 3. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. Caused PAPR flapping is only 0.2586 dB, and amaximum PAPR value is 4.2136. Both a PAPR value of an existing 4×HE-LTFsymbol and a PAPR value of an existing 2×HE-LTF symbol are greater than5 dB in the 20 MHz bandwidth.

TABLE 3 Phase Difference 20 MHz   1 4.1121 −1 3.9572 exp (−jπ/3) 4.2136exp (−j2π/3) 3.9550 PAPR_(max) − PAPR_(min) 0.2586

A suboptimal sequence is: HE-LTF₆₀(−120:4:120)={+1, −1, −1, BB_LTF_L,−1, BB_LTF_L, 0, BB_LTF_R, −1−1×BB_LTF_R, +1, +1, −1}, or may berepresented as

HE-LTF₆₀(−120:4:120) = {+1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, 0, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.0821 dB.

Referring to Table 4, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 4. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is 0.2398 dB, and a maximum PAPR value is4.3219 dB.

TABLE 4 Phase Difference 20 MHz   1 4.0821 −1 4.2189 exp (−jπ/3) 4.3219exp (−j2π/3) 4.1652 PAPRmax − PAPRmin 0.2398

Embodiment 2

Scenario: a subcarrier location B of a 1×HE-LTF in a 20 MHz bandwidth.

For ease of an interpolation operation in channel estimation, anothersubcarrier location pattern of an HE-LTF in a 1× mode in the 20 MHzbandwidth is −122:4:122. For example, additional ten subcarrier valuesare added based on sequences BB_LTF_L, BB_LTF_R, LTF_(left), andLTF_(right), to generate a 1×HE-LTF sequence. To ensure simpleimplementation, the ten subcarrier values are selected from {1, −1}. Anoptimal sequence is: HE-LTF₆₂(−122:4:122)={LTF_(right), −1, +1, −1, −1,+1, −1, −1, −1, +1, +1, LTF_(left)}, or may be represented as

HE-LTF₆₂(−122:4:122) = {+1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, −1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to +1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein, asdescribed above, −122:4:122 represents −122, −118, . . . , −6, −2, 2, 6,. . . , 118, and 122. In this case, corresponding pilot subcarrierlocations are ±22 and ±90, that is, there are four pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 3.7071 dB.

Referring to Table 5, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 5. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by an inter-stream phasedifference (caused by a matrix P, where the matrix P is defined inchapter 22.3.8.3.5 in the 11ac standard) between a data subcarrier and apilot subcarrier in the case of multiple spatial streams is only 0.2657,and a maximum PAPR value is 3.9728. Both a PAPR value of an existing4×HE-LTF symbol and a PAPR value of an existing 2×HE-LTF symbol aregreater than 5 dB in the 20 MHz bandwidth.

TABLE 5 Phase Difference 20 MHz  1 3.7071 −1 3.9149 exp(−jπ/3) 3.9728exp(−j2π/3) 3.8403 PAPRmax − PAPRmin 0.2657

A suboptimal sequence is: HE-LTF₆₂(−122:4:122)={BB_LTF_L, +1, +1, −1,−1×BB_LTF_L, −1, −1, +1, −1, −1×BB_LTF_R, +1, −1, −1, −1×BB_LTF_R}, ormay be represented as

HE-LTF₆₂(−122:4:122) = {+1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 3.8497 dB.

Referring to Table 6, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 6. PAPR flapping caused bythe inter-stream phase difference between a data subcarrier and a pilotsubcarrier in the case of multiple spatial streams is 0.4069, and amaximum PAPR value is 4.2566 dB.

TABLE 6 Phase Difference 20 MHz  1 3.8497 −1 4.2566 exp(−jπ/3) 4.1794exp(−j2π/3) 4.1750 PAPRmax − PAPRmin 0.4069

It should be noted that for Embodiment 2, in the subcarrier location Bscenario of the HE-LTF in the 1× mode in the 20 MHz bandwidth, a timedomain sequence obtained after an IFFT operation is performed on the1×HE-LTF sequence is LTF_(t)={LTF_(tq), −1×LTF_(tq), LTF_(tq),−1×LTF_(tq)}, where LTF_(tq) is first ¼ of the time domain sequence. Atransmit end may directly send an LTF_(tq) sequence Tx_LTF_(tq) to whicha cyclic prefix (CP, or referred to as GI) is added. It should be notedthat the CP sequence is a CP sequence obtained relative to an originalsequence (that is, the sequence LTF_(t)) that exists before truncation.If the transmit end uses 256-point IFFT, reference may be made to FIG.9A. FIG. 9A is a simple schematic diagram of a transmit end on asubcarrier location B of a 20 M 1×HE-LTF. Finally, a time windowingoperation and sending are performed.

In another equivalent solution, the transmit end may perform an IFFToperation on the 1×HE-LTF sequence to obtain a time domain sequence thatis LTF_(t)={LTF_(tq), −1 LTF_(tq), LTF_(tq), −1×LTF_(tq)}, whereLTF_(tq) is first ¼ of the time domain sequence. Then, the first ¼ istruncated to obtain the sequence LTF_(tq), and a CP of the LTF_(tq) isobtained for the sequence LTF_(tq) obtained by means of truncation.Then, after symbols of the CP sequence are negated (that is, all valuesin the CP are negated), the CP sequence is added before the LTF_(tq) toobtain a transmit sequence Tx_LTF_(tq). Finally, a time windowingoperation and sending are performed. If the transmit end uses 256-pointIFFT, reference may be made to FIG. 9B. FIG. 9B is a simple equivalentschematic diagram of a transmit end on a subcarrier location B of a 20 M1×HE-LTF.

In another equivalent solution, the transmit end may perform an IFFToperation on the 1×HE-LTF sequence to obtain a time domain sequence thatis LTF_(t)={LTF_(tq), −1×LTF_(tq), LTF_(tq), −1×LTF_(tq)}, whereLTF_(tq) is first ¼ of the time domain sequence. Then, a CP of theLTF_(t) is obtained for the sequence LTF_(t), and is added before theLTF_(t) to obtain the sequence LTF_(tp). Then, a CP of the sequenceLTF_(tp) and the first ¼ part of LTF_(t) are truncated (that is, the CPand the LTF_(tq)) to obtain a transmit sequence Tx_LTF_(tq). Finally, atime windowing operation and sending are performed. If the transmit enduses 256-point IFFT, reference may be made to FIG. 9C. FIG. 9C is asimple equivalent schematic diagram of a transmit end on a subcarrierlocation B of a 20 M 1×HE-LTF.

Correspondingly, it is assumed that a 1×HE-LTF time sequence received bya receive end is Rx_LTF_(tqr), and LTF_(tqr) is obtained after a CP isremoved. The receive end may first extend the time sequence toLTF_(tr)={LTF_(tqr), −1×LTF_(tqr), LTF_(tqr), −1×LTF_(tqr)}, and thenperform an FFT operation on the time sequence LTF_(tr). If the receiveend uses 256-point FFT, reference may be made to FIG. 10. FIG. 10 is asimple schematic diagram of a receive end on a subcarrier location B ofa 20 M 1×HE-LTF.

In FIG. 10, a time sequence received by a 1×HE-LTF part of the receiveend is Rx_LTF_(tq), and a sequence LTF_(tqr) is obtained after a frontCP is removed. Then, the LTF_(tqr) is repeated four times, and a symbolin the second time of repetition and a symbol in the fourth time ofrepetition are negated to obtain LTF_(tr)={LTF_(tqr), −1×LTF_(tqr),LTF_(tqr), −1×LTF_(tqr)}. Then, a 256-point FFT operation is performedon the LTF_(tr), to obtain a received 1×HE-LTF frequency domain sequencethat is referred to as an 1×Rx_HE-LTF.

Embodiment 3

Scenario: a 40 MHz bandwidth.

Additional 18 subcarrier values are added based on the following twogroups of sequences: LTF_(left) and LTF_(right), to generate a 1×HE-LTFsequence. To ensure simple implementation, the 18 subcarrier values areselected from {1, −1}.

For example, a sequence is: HE-LTF₁₂₂(−244:4:244)={LTF_(right), −1,LTF_(right), −1, −1, −1, +1, +1, −1, −1, −1, 0, +1, +1, +1, −1, −1, −1,−1, +1, −1×LTF_(left), +1, LTF_(left)}, or may be represented as

HE-LTF₁₂₂(244:4:244) = {+1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, −1, 0, +1, +1, +1, −1, −1, −1, −1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, 1 ischanged to −1, −1 is changed to 1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein,−244:4:244 represents −244, −240, . . . , −8, −4, 0, 4, 8, . . . , 240,and 244. In this case, corresponding pilot subcarrier locations are ±36,±104, ±144, and ±212, that is, there are eight pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.6555 dB.

Referring to Table 7, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 7. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is only 0.5273 dB, and a maximum PAPR valueis 4.6555 dB. In a worst case, both a PAPR value of an existing 4×HE-LTFsymbol and a PAPR value of an existing 2×HE-LTF symbol are greater than6 dB in the 40 MHz bandwidth.

TABLE 7 Phase Difference 20 MHz  1 4.6555 −1 4.1282 exp(−jπ/3) 4.5201exp(−j2π/3) 4.6117 PAPRmax − PAPRmin 0.5273

A suboptimal sequence is: HE-LTF₁₂₂(−244:4:244)={LTF_(right), −1, +1,+1, +1, −1, +1, −1, −1, +1, −1×LTF_(left), 0, −1×LTF_(right), +1, −1,−1, −1, −1, +1, +1, +1, +1, −1×LTF_(left)}, or may be represented as

HE-LTF₁₂₂(244:4:244) = {+1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, 0, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, +1, +1, +1, +1, −1, −1, +1, +1, −1, −1, +1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein,−244:4:244 represents −244, −240, . . . , −8, −4, 0, 4, 8, . . . , 240,and 244. In this case, corresponding pilot subcarrier locations are ±36,±104, ±144, and ±212, that is, there are eight pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.6831 dB.

Referring to Table 8, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 8. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is only 0.3397 dB, and a maximum PAPR valueis 4.8335 dB. In a worst case, both a PAPR value of an existing 4×HE-LTFsymbol and a PAPR value of an existing 2×HE-LTF symbol are greater than6 dB in the 40 MHz bandwidth.

TABLE 8 Phase Difference 20 MHz  1 4.6831 −1 4.4938 exp(−jπ/3) 4.7504exp(−j2π/3) 4.8335 PAPRmax − PAPRmin 0.3397

A further suboptimal sequence is: HE-LTF₁22(−244:4:244)={+1, +1, +1,LTF_(left), +1, LTF_(right), +1, −1, −1, +1, −1, 0, +1, −1×LTF_(left),−1, −1×LTF_(right), −1, −1, +1, +1, −1, +1, −1}, or may be representedas

HE-LTF₁₂₂(244:4:244) = {+1, +1, +1, +1, −1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, −1, 0, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, +1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein,−244:4:244 represents −244, −240, . . . , −8, −4, 0, 4, 8, . . . , 240,and 244. In this case, corresponding pilot subcarrier locations are ±36,±104, ±144, and ±212, that is, there are eight pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 5.1511 dB.

Referring to Table 9, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 9. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is only 0.1 dB, and a maximum PAPR value is5.1511 dB. In a worst case, both a PAPR value of an existing 4×HE-LTFsymbol and a PAPR value of an existing 2×HE-LTF symbol are greater than6 dB in the 40 MHz bandwidth.

TABLE 9 Phase Difference 20 MHz  1 5.1511 −1 5.0511 exp(−jπ/3) 5.0733exp(−j2π/3) 5.0643 PAPRmax − PAPRmin 0.1000

A still suboptimal sequence is: HE-LTF₁₂₂(−244:4:244)={+1, +1, −1,LTF_(left), +1, LTF_(right), +1, +1, −1, +1, +1, 0, −1, −1×LTF_(left),−1, −1×LTF_(right), −1, +1, +1, +1, +1, +1, −1}, or may be representedas

HE-LTF₁₂₂(244:4:244) = {+1, +1, −1, +1, −1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, +1, +1, 0, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, +1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.9848 dB.

Referring to Table 10, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 10. The phase difference iscaused by a matrix A, and the matrix A is defined in chapter 22.3.8.3.5in the 11ac standard. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is 0.3083 dB, and a maximum PAPR value is5.2026 dB.

TABLE 10 Phase Difference 20 MHz  1 4.9848 −1 4.8943 exp(−jπ/3) 5.0471exp(−j2π/3) 5.2026 PAPRmax − PAPRmin 0.3083

Embodiment 4

Scenario: an 80 MHz bandwidth.

Additional 42 subcarrier values are added based on the following twogroups of sequences: LTF_(left) and LTF_(right), to generate a 1×HE-LTFsequence. To ensure simple implementation, the 42 subcarrier values areselected from {+1, −1}. An optimal sequence is:

HE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1 × LTF_(left), −1 × LTF_(right), +1, −1, −1, −1, −1, −1, −1, +1, LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, +0, −1, +1, +1, −1, −1, LTF_(left), LTF_(right), −1, +1, −1, −1, +1, −1, −1, +1, LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, −1, +1, +1},

or, it may be represented as

HE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, 0, −1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein,−500:4:500 represents −500, −496, . . . , −8, −4, 0, 4, 8, . . . , 496,and 500. In this case, corresponding pilot subcarrier locations are ±24,±92, ±400, and ±468, that is, there are eight pilot subcarriers. In acase of a single spatial stream, a PAPR value of a 1× HE-LTF symbolgenerated according to the sequence is only 4.8609 dB. PAPR flappingcaused by an inter-stream phase difference between a data subcarrier anda pilot subcarrier in a case of multiple spatial streams is only 0.1413dB, and a maximum PAPR value is 5.0022 dB. In a worst case, both a PAPRvalue of an existing 4×HE-LTF symbol and a PAPR value of an existing2×HE-LTF symbol are greater than 6 dB in the 80 MHz bandwidth. It shouldbe noted herein that the optimal sequence herein means that left andright parts of the sequence can be combined to form a group of 160 M1×HE-LTF sequences with excellent performance.

The sequence in the foregoing implementation is a sequence representedby every four bits, and is expressed by 0 on a spacing location. Aperson of ordinary skill in the art may directly and undoubtedly obtaina 1×HE-LTF sequence, expressed in another manner, in the 80 M bandwidth.For example, a value 0 on another location is supplemented. A personskilled in the art may understand that the sequence is the same as theforegoing sequence in essence, and only a different expression manner isused and essence of the technical solution is not affected.

${HELTF}_{{- 500},300} = \begin{Bmatrix}{{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,} \\{0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,} \\{0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},} \\{0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,} \\{{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,} \\{0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,} \\{0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},} \\{0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,} \\{{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,} \\{0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,} \\{0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},} \\{0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,} \\{{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,} \\{0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,} \\{0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},} \\{0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,} \\{{- 1},0,0,0,0,0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,} \\{{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,} \\{0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,} \\{0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},} \\{0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,} \\{{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},0,0,} \\{0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,} \\{0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},} \\{0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,} \\{{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,} \\{0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,} \\{0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},} \\{0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,} \\{{- 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,} \\{0,{+ 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,} \\{0,0,{- 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{+ 1},0,0,0,{- 1},0,0,0,{- 1},0,0,0,{- 1},} \\{0,0,0,{+ 1},0,0,0,{+ 1}}\end{Bmatrix}$

Phase Difference 20 MHz  1 4.8609 −1 4.9858 exp(−jπ/3) 5.0022exp(−j2π/3) 5.0021 PAPRmax − PAPRmin 0.1413

A suboptimal sequence:

HE-LTF₂₅₀(−500:4:500) = {+1, −1, −1, +1, −1, +1, +1, −1, LTF_(left), LTF_(right), +1, +1, −1, +1, +1, −1, −1, −1, LTF_(left), −1 × LTF_(right), +1, −1, −1, −1, −1, 0, +1, +1, +1, −1, −1, LTF_(left), LTF_(right), −1, +1, +1, +1, +1, −1, +1, −1, −1 × LTF_(left), LTF_(right), +1, +1, +1, +1, −1, −1, −1, +1},

or may be represented as

HE-LTF₂₅₀(−500:4:500) = {+1, −1, −1, +1, −1, +1, +1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, 0, +1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, +1, +1, −1, −1, −1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers. Herein,−500:4:500 represents −500, −496, . . . , −8, −4, 0, 4, 8, . . . , 496,and 500. In this case, corresponding pilot subcarrier locations are ±24,±92, ±400, and ±468, that is, there are eight pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.8024 dB.

Referring to Table 11, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 11. PAPR flapping caused bythe inter-stream phase difference between a data subcarrier and a pilotsubcarrier in the case of multiple spatial streams is only 0.1324 dB,and a maximum PAPR value is 4.9348 dB. In a worst case, both a PAPRvalue of an existing 4×HE-LTF symbol and a PAPR value of an existing 2×HE-LTF symbol are greater than 6 dB in the 80 MHz bandwidth.

TABLE 11 Phase Difference 20 MHz  1 4.8024 −1 4.8680 exp(−jπ/3) 4.8809exp(−j2π/3) 4.9348 PAPRmax − PAPRmin 0.1324

A further suboptimal sequence is:

HE-LTF₂₅₀(−500:4:500) = {−1, +1, +1, +1, −1, +1, +1, +1, −1 × LTF_(left), −1 × LTF_(right), +1, −1, −1, −1, −1, −1, +1, +1, LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, +0, +1, −1, +1, +1, +1, −1 × LTF_(left), −1 × LTF_(right), +1, −1, +1, +1, −1, −1, +1, −1, −1 × LTF_(left), LTF_(right), −1, +1, +1, +1, +1, −1, −1, −1},

or may be represented as

HE-LTF₂₅₀(−500:4:500) = {−1, +1, +1, +1, −1, +1, +1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, +0, +1, −1, +1, +1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, +1, −1, −1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, +1, +1, −1, −1, −1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.97 dB.

Referring to Table 12, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 12. PAPR flapping caused bythe inter-stream phase difference between a data subcarrier and a pilotsubcarrier in the case of multiple spatial streams is only 0.26 dB, anda maximum PAPR value is 4.97 dB.

TABLE 12 Phase Difference 20 MHz  1 4.97 −1 4.71 exp(−jπ/3) 4.96exp(−j2π/3) 4.86 PAPRmax − PAPRmin 0.26

A still suboptimal sequence is:

HE-LTF₂₅₀(−500:4:500) = {−1, −1, −1, +1, +1, +1, +1, +1, −1 × LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, −1, −1, −1, LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, +0, −1, +1, +1, −1, −1, LTF_(left), LTF_(right), −1, +1, −1, −1, +1, −1, −1, +1, LTF_(left), −1 × LTF_(right), +1, −1, +1, −1, −1, −1, +1, +1},

or may be represented as

HE-LTF₂₅₀(−500:4:500) = {−1, −1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, +0, −1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1}.

In addition, the sequence also includes a sequence obtained afterpolarity of each value in the sequence is reversed (that is, +1 ischanged to −1, −1 is changed to +1, and 0 remains unchanged), andremaining subcarriers are 0, that is, empty subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 4.53 dB.

Referring to Table 13, a PAPR value caused by an inter-stream phasedifference between a data subcarrier and a pilot subcarrier in a case ofmultiple spatial streams is listed in Table 13. PAPR flapping caused bythe inter-stream phase difference between a data subcarrier and a pilotsubcarrier in the case of multiple spatial streams is only 0.52 dB, anda maximum PAPR value is 5.05 dB.

TABLE 13 Phase Difference 20 MHz  1 4.53 −1 4.91 exp(−jπ/3) 5.03exp(−j2π/3) 5.05 PAPRmax − PAPRmin 0.52

Embodiment 5

Solution 1 in a 160 MHz bandwidth.

A subcarrier of the 160 MHz bandwidth may be obtained by splicing two 80MHz subcarriers. A primary 80 M frequency band and a secondary 80 Mfrequency band may be spliced continuously or spaced by a particularbandwidth (for example, spaced by 100 MHz). In addition, frequency bandlocations of the primary 80 M frequency band and the secondary 80 Mfrequency band may be flexibly adjusted according to an actualsituation. Therefore, 1×HE-LTF sequences of the primary 80 M frequencyband and the secondary 80 M frequency band may be separately defined,and polarity is adjusted by using an entire 80 M sequence as a unit andbased on a spacing between the primary 80 M frequency band and thesecondary 80 M frequency band and frequency band order of the primary 80M frequency band and the secondary 80 M frequency band, to obtain alower PAPR.

Herein, it is assumed that the optimal sequence in Embodiment 4corresponds to HE-LTF_(80M) _(_) _(A), and HE-LTF_(80M) _(_)_(A)(−500:4:500)={L-LTF_(80M) _(_) _(A), 0, R-LTF_(80M) _(_) _(A)}. Thesequences L-LTF_(80M) _(_) _(A) and R-LTF_(80M) _(_) _(A) are used asbasic sequences, to respectively generate a primary 80 M sequence and asecondary 80 M sequence. A primary 80 M 1×HE-LTF sequence is LTF_(80M)_(_) _(Primary)={L-LTF_(80M) _(_) _(A), 0, R-LTF_(80M) _(_) _(A)}, and asecondary 80 M 1×HE-LTF sequence is LTF_(80M) _(_)_(Secondary)={L-LTF_(80M) _(_) _(A), 0, −1×R-LTF_(80M) _(_) _(A)}.

For ease of description, it is assumed that P1 indicates a polarityadjustment coefficient of the primary 80 M sequence, and P2 indicates apolarity adjustment coefficient of the secondary 80 M sequence. If P1 is+1, P2 may be +1 or −1. In this case, when a location relationship oftwo 80 M channels is [Primary 80 M, Secondary 80 M], a 160 M sequenceis: HE-LTF₅₀₀=[P1×LTF_(80M) _(_) _(Primary), BI, P2×LTF_(80M) _(_)_(Secondary)]. When a location relationship of two 80 M channels is[Secondary 80 M, Primary 80 M], a 160 M sequence is:HE-LTF₅₀₀=[P2×LTF_(80M) _(_) _(Secondary), BI, P1×LTF_(80M) _(_)_(Primary)]. The BI is a frequency spacing between subcarriers on edgesof the two 80 M channels (that is, the BI is a sequence carried on asubcarrier between the subcarriers on the edges of the two 80 Mchannels). When the primary 80 M channel and the secondary 80 M channelare adjacent, the BI={0, 0, 0, 0, 0}. When the primary 80 M channel andthe secondary 80 M channel are not adjacent, the BI may becorrespondingly adjusted. In addition, the primary 80 M channel and thesecondary 80 M channel may be independently generated, and then splicedto generate a 160 M frequency band.

Polarity adjustment coefficients of a primary 80 MHz bandwidth and asecondary 80 MHz bandwidth in two types of frequency band order andvarious frequency spacings are shown in the following table. Aprimary-secondary channel spacing is a center frequency spacing betweentwo 80 M frequency bands (a spacing of 80 MHz is obtained by splicingtwo adjacent 80 M channels). Specifically, corresponding PAPR values invarious cases are also shown in the table. The PAPR value is a maximumvalue between data and a pilot in four phase differences. It can belearned from the following table that there are only a few cases inwhich polarity of the primary 80 M sequence and polarity of thesecondary 80 M sequence need to be adjusted, and in most cases, theprimary 80 M sequence and the secondary 80 M sequence are directlyspliced. For example, when a location relationship of two adjacent 80 Mchannels is [Primary 80 M, Secondary 80 M], a 160 M sequence isspecifically HE-LTF₅₀₀(−1012:4:1012)={L-LTF_(80M) _(_) _(A), 0,R-LTF_(80M) _(_) _(A), 0, 0, 0, 0, 0, L-LTF_(80M) _(_) _(A), 0,−1×R-LTF_(80M) _(_) _(A)}.

Primary-secondary [Primary 80M, [Secondary 80M, channel spacingSecondary 80M] PAPR Primary 80M] PAPR (MHz) [P1, P2] (dB) [P2, P1] (dB)80 [+1, +1] 5.12 [+1, +1] 5.14 (adjacent) 100 [+1, +1] 5.15 [+1, +1]5.32 120 [+1, +1] 5.29 [+1, +1] 5.41 140 [+1, +1] 5.24 [+1, +1] 5.37 160[+1, +1] 5.30 [+1, +1] 5.32 180 [+1, +1] 5.33 [+1, +1] 5.40 200 [+1, +1]5.41 [+1, +1] 5.40 220 [+1, +1] 5.40 [−1, +1] 5.40 240 [+1, +1] 5.43[+1, +1] 5.42 >240 [+1, −1] ~5.44 [−1, +1] ~5.35

In addition, to reduce system implementation complexity, specific PAPRperformance may be selected for sacrifice. In various cases, the primary80 M sequence and the secondary 80 M sequence are directly spliced, toobtain a 1×HE-LTF sequence in the 160 M bandwidth.

The sequence in the foregoing implementation is a sequence representedby every four bits, and is 0 on a spacing location. The foregoingexample in which HE-LTF₅₀₀=[P1×LTF_(80M) _(_) _(Primary), BI,P2×LTF_(80M) _(_) _(Secondary)], P1 is +1, and P2 is +1 is used. Aperson of ordinary skill in the art may directly and undoubtedly obtaina sequence expressed in another manner, that is, a manner ofsupplementing a value 0 on another location in the entire sequence. Aperson skilled in the art may understand that the sequence issubstantially the same as the foregoing sequence, and only a differentexpression manner is used and essence of the technical solution is notaffected.

HE-LTF_(−1012:1:1012)={LTF′_(80M) _(_)_(Primary),0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,LTF′_(80M) _(_)_(Secondary)}, where

LTF′_(80M) _(_) _(Primary) ={L-LTF′_(80M) _(_) _(A),0,R-LTF′_(80M) _(_)_(A)}, and

LTF′_(80M) _(_) _(Secondary) ={L-LTF′_(80M) _(_) _(A),0,−1×R-LTF′_(80M)_(_) _(A)}.

It can be directly and undoubtedly learned from the sequence inEmbodiment 4 that

L-LTF_(80 M_A)^(′) = {−1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0}, andR-LTF_(80 M_A)^(′) = {0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, −1, 0, 0, 0, +1, 0, 0, 0, +1}.

Solution 2 in a 160 MHz bandwidth:

A subcarrier of the 160 MHz bandwidth is obtained by repeating an 80 MHzsubcarrier and then directly splicing 80 MHz subcarriers. Therefore, aIx HE-LTF sequence in the 160 M bandwidth is generated based on thesuboptimal Ix HE-LTF sequence in the 80 M bandwidth in Embodiment 4. Forease of description, the suboptimal sequence is referred to as anHE-LTF_(80M), and HE-LTF_(80M)(−500:4:500)={L-LTF_(80M), 0,R-LTF_(80M)}. A sequence in the solution 1 in the 160 MHz bandwidth is:HE-LTF₅₀₀(−1012:4:1012)={L-LTF_(80M), 0, R-LTF_(80M), 0, 0, 0, 0, 0,−1×L-LTF_(80M), 0, R-LTF_(80M)}, and remaining subcarriers are 0, thatis, empty subcarriers. Herein, −1012:4:1012 represents −1012, −1008, . .. , −8, −4, 0, 4, 8, . . . , 1008, and 1012. In this case, correspondingpilot subcarrier locations are ±44, ±112, ±420, ±488, ±536, ±604, ±912,and ±980, that is, there are 16 pilot subcarriers.

In a case of a single spatial stream, a PAPR value of a 1×HE-LTF symbolgenerated according to the sequence is only 5.7413 dB.

A PAPR value caused by an inter-stream phase difference between a datasubcarrier and a pilot subcarrier in a case of multiple spatial streamsis listed in Table 14. PAPR flapping caused by the inter-stream phasedifference between a data subcarrier and a pilot subcarrier in the caseof multiple spatial streams is 0.3948 dB, and a maximum PAPR value isonly 5.9667 dB.

TABLE 14 Phase Difference 20 MHz  1 5.7413 −1 5.5883 exp(−jπ/3) 5.9485exp(−j2π/3) 5.9667 PAPRmax − PAPRmin 0.2254

Another solution in a 160 MHz bandwidth:

A subcarrier of the 160 MHz bandwidth may be obtained by splicing two 80MHz subcarriers. A primary 80 M frequency band and a secondary 80 Mfrequency band may be spliced continuously or spaced by a particularbandwidth (for example, spaced by 100 MHz). In addition, frequency bandlocations of the primary 80 M frequency band and the secondary 80 Mfrequency band may be flexibly adjusted according to an actualsituation. Therefore, 1×HE-LTF sequences of the primary 80 M frequencyband and the secondary 80 M frequency band may be separately defined,and polarity is adjusted by using an entire 80 M sequence as a unit andbased on a spacing between the primary 80 M frequency band and thesecondary 80 M frequency band and frequency band order of the primary 80M frequency band and the secondary 80 M frequency band, to obtain alower PAPR.

Herein, the suboptimal sequence and the further suboptimal sequence inEmbodiment 4 are respectively used as a primary 80 M sequence and asecondary 80 M sequence, and are spliced to obtain a new 1×HE-LTFsequence in the 160 MHz bandwidth.

For ease of description, the suboptimal sequence in Embodiment 4 isreferred to as LTF_(80M) _(_) _(Primary), and the further suboptimalsequence in Embodiment 4 is referred to as LTF_(80M) _(_) _(Secondary).It is assumed that P1 indicates a polarity adjustment coefficient of theprimary 80 M sequence, and P2 indicates a polarity adjustmentcoefficient of the secondary 80 M sequence. If P1 is +1, P2 may be +1 or−1. In this case, when a placement relationship of two 80 M channels is[Primary 80 M, Secondary 80 M], a 160 M sequence is:HE-LTF₅₀₀=[P1×LTF_(80M) _(_) _(Primary), BI, P2×LTF_(80M) _(_)_(Secondary)]. When a placement relationship of two 80 M channels is[Secondary 80 M, Primary 80 M], a 160 M sequence is:HE-LTF₅₀₀=[P2×LTF_(80M) _(_) _(Secondary), BI, P1×LTF_(80M) _(_)_(Primary)]. The BI is a frequency spacing between subcarriers on edgesof two 80 M channels. When the primary 80 M channel and the secondary 80M channel are adjacent, the BI={0, 0, 0, 0, 0}. When the primary 80 Mchannel and the secondary 80 M channel are not adjacent, the BI may becorrespondingly adjusted. In addition, the primary 80 M channel and thesecondary 80 M channel may be independently generated, and then splicedto generate a 160 M frequency band.

Polarity adjustment coefficients of a primary 80 MHz bandwidth and asecondary 80 MHz bandwidth in two types of frequency band order andvarious frequency spacings are shown in the following Table 15. Aprimary-secondary channel spacing is a center frequency spacing betweentwo 80 M frequency bands (a spacing of 80 MHz is obtained by splicingtwo adjacent 80 M channels).

Specifically, corresponding PAPR values in various cases are also shownin Table 15. The PAPR value is a maximum value between data and a pilotin four phase differences. It can be learned from the following tablethat there are only few cases in which polarity of the primary 80 Msequence and polarity of the secondary 80 M sequence need to beadjusted, and in most cases, the primary 80 M sequence and the secondary80 M sequence are directly spliced.

TABLE 15 Primary-secondary [Primary 80M, [Secondary 80M, channel spacingSecondary 80M] PAPR Primary 80M] PAPR (MHz) [P1, P2] (dB) [P2, P1] (dB)80 [+1, +1] 5.48 [+1, +1] 5.59 (adjacent) 100 [+1, +1] 5.48 [+1, +1]5.51 120 [+1, −1] 5.58 [+1, +1] 5.58 140 [+1, +1] 5.51 [+1, +1] 5.47 160[+1, +1] 5.63 [−1, +1] 5.49 180 [+1, −1] 5.53 [+1, +1] 5.65 200 [+1, −1]5.61 [+1, +1] 5.54 220 [+1, +1] 5.51 [−1, +1] 5.51 240 [+1, +1] 5.59[+1, +1] 5.60 >240 [+1, +1] ~5.63 [+1, +1] ~5.57

In addition, to reduce system implementation complexity, specific PAPRperformance may be selected for sacrifice. In various cases, the primary80 M sequence and the secondary 80 M sequence are directly spliced, toobtain a 1×HE-LTF sequence in the 160 M bandwidth.

In the embodiments, all the 1×HE-LTF sequences are characterized by agood PAPR in different bandwidths, and a PAPR is characterized byextremely small fluctuation in the case of multiple spatial streams, sothat a power amplifier can be effectively used, and power can be betterenhanced in a long distance transmission mode to adapt to longerdistance transmission.

The present embodiments may be applied to a wireless local area networkthat includes but is not limited to a Wi-Fi system represented by802.11a, 802.11b, 802.11g, 802.11n, or 802.11ac; or may be applied to anext-generation Wi-Fi system or a next-generation wireless local areanetwork system.

It further provides a data transmission apparatus that may perform theforegoing method. FIG. 11 is an example (for example, some components inthe figure such as an access point, a station, and a chip are optional)of a schematic structural diagram of a data transmission apparatus inthe embodiments. As shown in FIG. 11, a data transmission apparatus 1200may be implemented by using a bus 1201 as a general bus architecture.The bus 1201 may include any quantity of interconnected buses andbridges according to specific application and an overall designconstraint condition that are of the data transmission apparatus 1200.Various circuits are connected together by using the bus 1201. Thesecircuits include a processor 1202, a storage medium 1203, and a businterface 1204. In the data transmission apparatus 1200, a networkadapter 1205 and the like are connected via the bus 1201 by using thebus interface 1204. The network adapter 1205 may be configured to:implement a signal processing function at a physical layer in a wirelesslocal area network, and send and receive a radio frequency signal byusing an antenna 1207. A user interface 1206 may be connected to a userterminal such as a keyboard, a display, a mouse, or a joystick. The bus1201 may be further connected to various other circuits, such as atiming source, a peripheral device, a voltage regulator, and a powermanagement circuit. These circuits are known in the art. Therefore,details are not described.

Alternatively, the data transmission apparatus 1200 may be configured asa general-purpose processing system. The general-purpose processingsystem includes: one or more microprocessors that provide a processorfunction, and an external memory that provides at least one part of thestorage medium 1203. All the components are connected to another supportcircuit by using an external bus architecture.

Alternatively, the data transmission apparatus 1200 may be implementedby using an ASIC (application-specific integrated circuit) that includesthe processor 1202, the bus interface 1204, and the user interface 1206,and at least one part that is of the storage media 1203 and that isintegrated into a single chip. Alternatively, the data transmissionapparatus 1200 may be implemented by using one or more FPGAs (fieldprogrammable gate array), a PLD (programmable logic device), acontroller, a state machine, gate logic, a discrete hardware component,any other appropriate circuit, or any combination of circuits that canperform various functions described in the present embodiments.

The processor 1202 is responsible for bus management and generalprocessing (including executing software stored on the storage medium1203). The processor 1202 may be implemented by using one or moregeneral-purpose processors and/or dedicated processors. The processorincludes, for example, a microprocessor, a microcontroller, a DSPprocessor, or another circuit that can execute software. Regardless ofwhether the software is referred to as software, firmware, middleware,micro code, hardware description language, or the like, the softwareshould be broadly construed as an instruction, data, or any combinationthereof.

It is shown in FIG. 11 that the storage medium 1203 is separated fromthe processor 1202. However, a person skilled in the art easilyunderstands that the storage medium 1203 or any part of the storagemedium 1203 may be located outside the data transmission apparatus 1200.For example, the storage medium 1203 may include a transmission line, acarrier waveform obtained by means of data modulation, and/or a computerproduct separated from a wireless node. All the media may be accessed bythe processor 1202 by using the bus interface 1204. Alternatively, thestorage medium 1203 or any part of the storage medium 1203 may beintegrated into the processor 1202, for example, may be a cache and/or ageneral-purpose register.

The processor 1202 may perform the foregoing embodiment, and details arenot described herein.

A person of ordinary skill in the art may understand that all or some ofthe steps of the method embodiments may be implemented by a programinstructing relevant hardware. The program may be stored in a computerreadable storage medium. When the program runs, the steps of the methodembodiments are performed. The foregoing storage medium includes: anymedium that can store program code, such as a ROM, a RAM, a magneticdisk, or an optical disc.

What is claimed is:
 1. A channel estimation information transmissionmethod comprising: determining, by a transmit apparatus, a highefficiency long training field (HE-LTF) sequence in frequency domainaccording to a transmission bandwidth and a mode of a HE-LTF field;sending, by the transmit apparatus, a time-domain signal, according to anumber of orthogonal frequency division multiplexing (OFDM) symbols ofthe HE-LTF field, N_(HELTF), and the determined HE-LTF sequence in thefrequency domain; wherein the HE-LTF sequence in the frequency domain ina 1×HE-LTF mode over a 160 MHz bandwidth corresponds to:HE-LTF=[P1×LTF_(80 MHz) _(_) _(Primary) ,BI,P2×LTF_(80 MHz) _(_)_(Secondary)], wherein P1 is +1, P2 is +1 or −1,LTF_(80 MHz) _(_) _(Primary) ={L-LTF_(80 MHz) _(_) _(A),0,R-LTF_(80 MHz)_(_) _(A)},LTF_(80 MHz) _(_) _(Secondary) ={L-LTF_(80 MHz) _(_)_(A),0,−1×R-LTF_(80 MHz) _(_) _(A)} and the BI is a sequence carried ona subcarrier between subcarriers on edges of two 80 M channels.
 2. Themethod according to claim 1, wherein {L-LTF_(80 MHz) _(_) _(A), 0,R-LTF_(80 MHz) _(_) _(A)}=HE-LTF₂₅₀(−500:4:500), andHE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, −1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, 0, −1, −1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, −1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1},wherein each value of the HE-LTF₂₅₀(−500:4:500) corresponds to eachvalue on subcarrier with indexes −500, −496, . . . , −8, −4, 0, 4, 8, .. . , 496, and 500, respectively, and values on remaining subcarriersare
 0. 3. The method according to claim 2, wherein the primary 80 MHz(LTF_(80 MHz) _(_) _(Primary)) channel and the secondary 80 MHz(LTF_(80 MHz) _(_) _(Secondary)) channel in the 160 MHz bandwidth areadjacent, the BI={0, 0, 0, 0, 0}.
 4. The method according to claim 1,the method further comprising: before determining the HE-LTF sequence infrequency domain, receiving a trigger frame to indicate uplinkscheduling information in an uplink multi-user multiple-inputmultiple-output (UL-MU-MIMO) transmission, wherein the uplink schedulinginformation includes the transmission bandwidth and the number of HE-LTFsymbols.
 5. The method according to claim 1, wherein the method furthercomprising: determining, the N_(HELTF), based on a total number ofspace-time streams, N_(STS), in a single user transmission, or, adownlink multi-user multiple-input multiple-output (DL-MU-MIMO)transmission, wherein the N_(HELTF) and the N_(STS) correspond to:N_(STS) N_(HELTF) 1 1 2 2 3 4 4 4 5 6 6 6 7 8 8 8


6. A channel estimation information process method, comprising:receiving, by a receive apparatus, a preamble which comprises a highefficiency long training field (HE-LTF); obtaining, by the receiveapparatus, a channel estimation value of a corresponding subcarrierlocation, according to the received HE-LTF field and a HE-LTF sequencein frequency domain; wherein the HE-LTF sequence in the frequency domainin a 1×HE-LTF mode over a 160 MHz bandwidth corresponds to:HE-LTF=[P1×LTF_(80 MHz) _(_) _(Primary) ,BI,P2×LTF_(80 MHz) _(_)_(Secondary)], wherein P1 is +1, P2is +1 or −1,LTF_(80 MHz) _(_) _(Primary) ={L-LTF_(80 MHz) _(_) _(A),0,R-LTF_(80 MHz)_(_) _(A)},LTF_(80 MHz) _(_) _(Secondary) ={L-LTF_(80 MHz) _(_)_(A),0,−1×R-LTF_(80 MHz) _(_) _(A)}, and the BI is a frequency spacingbetween subcarriers on edges of two 80 MHz channels.
 7. The methodaccording to claim 6, wherein {L-LTF_(80 MHz) _(_) _(A), 0,R-LTF_(80 MHz) _(_) _(A)}=HE-LTF₂₅₀(−500:4:500), andHE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, 0, −1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1},wherein each value of the HE-LTF₂₅₀(−500:4:500) corresponds to eachvalue on subcarrier with indexes −500, −496, . . . , −8, −4, 0, 4, 8, .. . , 496, and 500, respectively, and values on remaining subcarriersare
 0. 8. The method according to claim 7, wherein when a primary 80 MHz(LTF_(80 MHz) _(_) _(Primary)) channel and a secondary 80 MHz(LTF_(80 MHz) _(_) _(Secondary)) channel in the 160 MHz bandwidth areadjacent, the BI={0, 0, 0, 0, 0}.
 9. The method according to claim 6,further comprising: before receiving the preamble which comprises theHE-LTF field, sending, by the receive apparatus, a trigger frame toindicate uplink scheduling information in an uplink multi-usermultiple-input multiple-output (UL-MU-MIMO) transmission, wherein theuplink scheduling information includes a transmission bandwidth and anumber of HE-LTF symbols.
 10. The method according to claim 6, furthercomprising: obtaining, by the receive apparatus, a transmissionbandwidth BW, a total number of space-time streams, N_(STS), and a modeof a high efficiency long training HE-LTF field, according toinformation carried in a signal field in the preamble, in a single usertransmission, or, a downlink multi-user multiple-input multiple-output(DL-MU-MIMO) transmission; determining, by the receive apparatus, anumber of orthogonal frequency division multiplexing (OFDM) symbols ofthe HE-LTF field, N_(HELTF), based on the total number of space-timestreams, N_(STS); and determining, by the receive apparatus, the HE-LTFsequence in frequency domain according to the transmission bandwidth andthe mode of the HE-LTF field.
 11. A channel estimation informationtransmission apparatus, comprising: a memory storage comprisinginstructions; and one or more processors in communication with thememory, wherein the one or more processors execute the instructions to:determine a high efficiency long training field HE-LTF sequence infrequency domain according to a transmission bandwidth and a mode of aHE-LTF field; send a time-domain signal, according to a number oforthogonal frequency division multiplexing (OFDM) symbols of the HE-LTFfield, N_(HELTF), and the determined HE-LTF sequence in the frequencydomain; wherein the HE-LTF sequence in the frequency domain in a1×HE-LTF mode over a 160 MHz bandwidth corresponds to:HE-LTF=[P1×LTF_(80 MHz) _(_) _(Primary) ,BI,P2×LTF_(80 MHz) _(_)_(Secondary)], wherein P1 is +1, P2 is +1 or −1,LTF_(80 MHz) _(_) _(Primary) ={L-LTF_(80 MHz) _(_) _(A),0,R-LTF_(80 MHz)_(_) _(A)},LTF_(80 MHz) _(_) _(Secondary) ={L-LTF_(80 MHz) _(_)_(A),0,−1×R-LTF_(80 MHz) _(_) _(A)} and the BI is a sequence carried ona subcarrier between subcarriers on edges of two 80 M channels.
 12. Theapparatus according to claim 11, wherein {L-LTF_(80 MHz) _(_) _(A), 0,R-LTF_(80 MHz) _(_) _(A)}=HE-LTF₂₅₀(−500:4:500), andHE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, + 1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, 0, −1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, − 1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, +1, + 1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1},wherein each value of the HE-LTF₂₅₀(−500:4:500) corresponds to eachvalue on subcarrier with indexes −500, −496, . . . , −8, −4, 0, 4, 8, .. . , 496, and 500, respectively, and values on remaining subcarriersare
 0. 13. The apparatus according to claim 12, wherein the primary 80MHz (LTF_(80 MHz) _(_) _(Primary)) channel and the secondary 80 MHz(LTF_(80 MHz) _(_) _(Secondary)) channel in the 160 MHz bandwidth areadjacent, the BI={0, 0, 0, 0, 0}.
 14. The apparatus according to claim11, wherein the one or more processors execute the instructions furtherto: transmit apparatus before determine the HE-LTF sequence in thefrequency domain, receive a trigger frame to indicate uplink schedulinginformation in an uplink multi-user multiple-input multiple-output(UL-MU-MIMO) transmission, wherein the uplink scheduling informationincludes the transmission bandwidth and the number of HE-LTF symbols.15. The apparatus according to claim 12, wherein the one or moreprocessors execute the instructions further to: transmit apparatusdetermine, the N_(HELTF), based on a total number of space-time streams,N_(STS), in a single user transmission, or, in a downlink multi-usermultiple-input multiple-output (DL-MU-MIMO) transmission, wherein theN_(HELTF) and the N_(STS) correspond to: N_(STS) N_(HELTF) 1 1 2 2 3 4 44 5 6 6 6 7 8 8 8


16. A channel estimation information process apparatus, comprising: amemory storage comprising instructions; and one or more processors incommunication with the memory, wherein the one or more processorsexecute the instructions to: receive a preamble which comprises a highefficiency long training field (HE-LTF); obtain a channel estimationvalue of a corresponding subcarrier location, according to the receivedHE-LTF field and a HE-LTF sequence in frequency domain; wherein theHE-LTF sequence in the frequency domain in a 1×HE-LTF mode over a 160MHz bandwidth corresponds to:HE-LTF=[P1×LTF_(80 MHz) _(_) _(Primary) ,BI,P2×LTF_(80 MHz) _(_)_(Secondary)], wherein P1 is +1, P2 is +1 or −1,LTF_(80 MHz) _(_) _(Primary) ={L-LTF_(80 MHz) _(_) _(A),0,R-LTF_(80 MHz)_(_) _(A)},LTF_(80 MHz) _(_) _(Secondary) ={L-LTF_(80 MHz) _(_)_(A),0,−1×R-LTF_(80 MHz) _(_) _(A)}, and the BI is a frequency spacingbetween subcarriers on edges of two 80 MHz channels.
 17. The apparatusaccording to claim 16, wherein {L-LTF_(80 MHz) _(_) _(A), 0,R-LTF_(80 MHz) _(_) _(A)}=HE-LTF₂₅₀(−500:4:500), andHE-LTF₂₅₀(−500:4:500) = {−1, −1, +1, +1, +1, +1, +1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, 0, −1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, −1, −1, +1, −1, −1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, +1, +1, +1, −1, +1, +1, −1, −1, +1, −1, +1, −1, +1, +1, +1, +1, +1, −1, −1, +1, +1, −1, +1, −1, +1, −1, −1, −1, −1, +1, −1, +1, −1, −1, −1, +1, +1},wherein each value of the HE-LTF₂₅₀(−500:4:500) corresponds to eachvalue on subcarrier with indexes −500, −496, . . . , −8, −4, 0, 4, 8, .. . , 496, and 500, respectively, and values on remaining subcarriersare
 0. 18. The apparatus according to claim 17, wherein when a primary80 MHz (LTF_(80 MHz) _(_) _(Primary)) channel and a secondary 80 MHz(LTF_(80 MHz) _(_) _(Secondary)) channel in the 160 MHz bandwidth areadjacent, the BI={0, 0, 0, 0, 0}.
 19. The apparatus according to claim16, wherein the one or more processors execute the instructions furtherto: receive apparatus before receive the preamble which comprises theHE-LTF field, send a trigger frame to indicate uplink schedulinginformation in an uplink multi-user multiple-input multiple-output(UL-MU-MIMO) transmission, wherein the uplink scheduling informationincludes a transmission bandwidth and a number of HE-LTF symbols. 20.The apparatus according to claim 16, wherein the one or more processorsexecute the instructions further to: obtain a transmission bandwidth BW,a total number of space-time streams, N_(STS), and a mode of a HE-LTFfield, according to information carried in a signal field in thepreamble, in a single user transmission, or, in a downlink multi-usermultiple-input multiple-output (DL-MU-MIMO) transmission; determine, anumber of orthogonal frequency division multiplexing (OFDM) symbols ofthe HE-LTF field, N_(HELTF), based on the total number of space-timestreams, N_(STS); and determine the HE-LTF sequence in frequency domainaccording to the transmission bandwidth and the mode of the HE-LTFfield.